D.2 Electric and magnetic fields

Draw the interaction

For two like point charges, draw arrows away from each other. For two unlike point charges, draw arrows toward each other. The direction is determined by the sign combination; the force magnitude also depends on charge magnitudes and separation, which Coulomb’s law quantifies.

Extend to a third charge

If a third charge is present, find the force from each other charge separately and add the force vectors. Do not decide the net direction by charge sign alone: compare the individual vectors and their magnitudes.

Worked example — unlike charges in air

For $q_1=4.5\times10^{-8}\,\mathrm{C}$, $q_2=-1.3\times10^{-7}\,\mathrm{C}$ and $r=3.2\times10^{-2}\,\mathrm{m}$, $F=(8.99\times10^9)|q_1q_2|/r^2=5.1\times10^{-2}\,\mathrm{N}$. The force is attractive because the charges have opposite signs.

Read the scaling

Doubling either charge doubles the force. Doubling the separation reduces the force to one quarter. If the medium has permittivity $\varepsilon$ rather than $\varepsilon_0$, use $k=1/(4\pi\varepsilon)$; greater permittivity reduces the force for the same charges and separation.

Choose the point-charge model

Spherical charged bodies can be treated as point charges at their centres when the geometry permits. Use centre-to-centre separation and convert charge units, such as microcoulombs, before substitution. For several charges, calculate each force vector and add them.

Use it at a junction

In a steady circuit, charge does not accumulate at a junction. The current entering equals the current leaving, for example $I_1=I_2+I_3$. This is a consequence of charge conservation, not a separate rule that overrides it.

Track the system boundary

When charge appears to change on one object, include the other object, the conductor or the ground in the system. Electrons may move across the chosen boundary, so the object’s charge changes while the total charge of the larger isolated system remains constant.

Read the evidence

Repeating the measurement for many drops gives charges that are integer multiples of a smallest value, the elementary charge $e$: $q=ne$, where $n$ is an integer. This pattern is evidence that electric charge is quantized rather than continuously variable.

Explain the method

The experiment varies the electric field until a drop is held stationary, then uses the known mass and field to infer its charge. It is the repeated integer-multiple pattern—not one isolated drop—that supports the quantization conclusion.

Transfer by induction

Bring a charged object near a conductor without touching it. Charges in the conductor separate by repulsion and attraction. If the conductor is connected to ground while the charged object remains nearby, electrons can flow to or from Earth. Disconnect the ground first, then remove the external charged object, leaving the conductor with a net charge.

Transfer by contact and grounding

Touching a charged conductor to another conductor allows charge to redistribute between them. Grounding connects an object to a very large charge reservoir: electrons can leave an object or enter it, depending on the nearby charge and the object’s potential.

Worked example — point-charge field

At $r=1.0\,\mathrm{m}$ from $Q=+2.9\times10^{-8}\,\mathrm{C}$, $E=(8.99\times10^9)(2.9\times10^{-8})/(1.0)^2=2.6\times10^2\,\mathrm{N\,C^{-1}}$. Because the source is positive, the field points radially outward.

Add fields as vectors

For more than one source, calculate each electric-field vector at the point and add them. Do not add magnitudes unless all field vectors point in the same direction. The force on a particular charge is then $F=qE$, with its direction reversed from the field if the charge itself is negative.

Read qualitative strength

Where field lines are closer together, the field is stronger; where they spread out, it is weaker. This is a qualitative representation unless the diagram specifies equal field-line intervals or a scale.

Connect density to distance

Around an isolated point charge, field lines spread as distance increases, so the field becomes weaker. A denser pattern near the charge is consistent with the inverse-square dependence of field strength. In a uniform field, equal spacing indicates constant field strength.

Check the representation

Density comparisons are meaningful only when the diagram uses the same line convention and potential/field intervals. Do not infer exact numerical values from arbitrary artwork; use labels or a scale if a calculation is required.

Worked example — required potential difference

For $E=1.0\times10^6\,\mathrm{V\,m^{-1}}$ and $d=0.50\,\mathrm{cm}=5.0\times10^{-3}\,\mathrm{m}$, $V=Ed=(1.0\times10^6)(5.0\times10^{-3})=5.0\times10^3\,\mathrm{V}$. The result applies to the uniform central region.

Read the direction

Electric field lines point from the positive plate to the negative plate, so a positive charge accelerates in that direction and a negative charge accelerates oppositely. The field strength can be expressed in $\mathrm{N\,C^{-1}}$ or equivalently $\mathrm{V\,m^{-1}}$.

Check the boundary

The formula assumes a uniform region and neglects edge effects. Use the perpendicular plate separation in metres; do not use the diagonal distance or the plate length.

Use the right-hand rule

Around a long straight current-carrying wire, the field lines are concentric circles centred on the wire. Point the right thumb in the conventional current direction; curled fingers give the magnetic-field direction. If electrons move into the page, conventional current is out of the page, so reverse the electron-motion direction before applying the rule.

Interpret energy transfer

Work done by an external agent changes the electric potential energy. If a positive charge moves toward a negative source, the electric field can do positive work while the potential energy decreases. If an electron is accelerated through a potential difference, the lost electric potential energy can become kinetic energy.

Track the system

For several charges, electric potential energy belongs to the whole charge configuration and is built from pair interactions. State the reference and the system before assigning a sign; do not confuse the energy of a charge configuration with electric potential at one point.

Worked example — two negative conductors

Radii $2.5\,\mathrm{cm}$ and $1.5\,\mathrm{cm}$, separated by a $1.7\,\mathrm{cm}$ surface gap, give $r=5.7\times10^{-2}\,\mathrm{m}$. For charges $-4.7\times10^{-8}\,\mathrm{C}$ and $-6.3\times10^{-8}\,\mathrm{C}$, $E_p=(8.99\times10^9)q_1q_2/r=+4.7\times10^{-4}\,\mathrm{J}$. Positive energy matches repulsion between like charges.

Interpret the sign

Like charges give positive potential energy because work is required to bring them together. Unlike charges give negative potential energy because the electric field releases energy as they approach. Increasing separation moves the energy toward zero.

Use changes in energy

When a charge configuration changes, calculate the final minus initial potential energy. The external work and work done by the electric field have opposite signs under a quasistatic convention. Use the actual separation between charge centres, not the physical radius of either object.

Add potentials algebraically

Electric potential is a scalar, so contributions from several point charges add with their signs:
$$V_e=\sum_i k\frac{Q_i}{r_i}$$
There is no vector-angle calculation when combining potential values.

Interpret the result

A positive source contributes positive potential and a negative source contributes negative potential. A point can have zero net potential because positive and negative contributions cancel, even though the electric field there is not necessarily zero.

Worked example — negative source

For $Q=-1.00\times10^{-8}\,\mathrm{C}$ at $r=1.00\,\mathrm{m}$, $V_e=(8.99\times10^9)(-1.00\times10^{-8})/(1.00)=-89.9\,\mathrm{V}$. At $2.00\,\mathrm{m}$, it is $-45.0\,\mathrm{V}$: farther away, the negative potential increases toward zero.

Combine sources

For several point charges, calculate each $kQ_i/r_i$ and add the scalar values. Use centre-to-point distance and convert all distances and charges before substitution. The potential does not depend on the test charge used to define it.

Check conducting spheres

Inside a charged conducting sphere in electrostatic equilibrium, the electric field is zero and the potential is constant throughout the interior. The potential need not be zero; it equals the surface potential for the ideal spherical case.

Worked example — tangent gradient

If a tangent changes from $-26\,\mathrm{kV}$ to $0\,\mathrm{V}$ over $8.0\,\mathrm{cm}=8.0\times10^{-2}\,\mathrm{m}$, $E=-[0-(-26\times10^3)]/(8.0\times10^{-2})=-3.3\times10^5\,\mathrm{V\,m^{-1}}$. The negative sign gives the field direction in the chosen coordinate.

Interpret the sign

The negative sign means the electric field points toward decreasing potential. A negative slope of $V_e$ against position corresponds to a positive field component in that coordinate direction; state whether the question wants a signed component or a magnitude.

Connect field to motion

A negative charge experiences force opposite to the electric field. Therefore its acceleration direction is opposite to the direction of decreasing potential, even though the field itself is always defined using a positive test charge.

Worked example — moving between equipotentials

A $+2.0\,\mathrm{C}$ charge moves from $40\,\mathrm{V}$ to $20\,\mathrm{V}$. Then $W_{\mathrm{on}}=(2.0)(20-40)=-40\,\mathrm{J}$. Its electric potential energy falls by $40\,\mathrm{J}$; if free, that energy can become kinetic energy.

Connect to kinetic energy

If only the electric field does work, $W_{\rm field}=-q\Delta V_e$, and the change in kinetic energy equals this work. A positive charge moving to lower potential can gain kinetic energy; a negative charge may gain kinetic energy moving to higher potential.

Use endpoints

Because electrostatic fields are conservative, the work between two points does not depend on the path. Motion along an equipotential has $\Delta V_e=0$ and therefore zero work by the field.

Read spacing carefully

If equal potential intervals are drawn, closer equipotential lines indicate a larger potential gradient and stronger electric field. Use labels and the drawing convention; arbitrary visual spacing alone does not provide a numerical field.

Apply it between plates

For oppositely charged parallel plates, field lines are approximately straight and perpendicular to the plates; equipotential surfaces are parallel to the plates. This is why the potential changes across the separation but remains constant along a plate.

Predict motion

A positive charge accelerates along the electric field, toward lower potential. A negative charge accelerates opposite to the field, toward higher potential. The field direction and particle-force direction must be kept separate.